Solution to Problem 3
We prove that the answer is
[ \boxed{c_n=\frac{2^n}{2^{n+1}-1}}. ]
Put
[ \delta=\frac1{2^{n+1}-1}. ]
Then
[ \frac{1+\delta}{2}=\frac{2^n}{2^{n+1}-1}. \tag{1} ]
1. A fact about the claiming stage
Suppose the final piece lengths, in nonincreasing order, are
[ x_1\ge x_2\ge\cdots\ge x_m>0. ]
If Liu always takes a longest remaining piece, then on his (j)-th move he receives a piece of length at least (x_{2j-1}). Indeed, before that move at most (2j-2) pieces have been removed, so at least one of the largest (2j-1) pieces remains.
Thus Liu can guarantee at least
[ x_1+x_3+x_5+\cdots. \tag{2} ]
Define the alternating gap
[ D=x_1-x_2+x_3-x_4+\cdots, ]
where a missing last even-indexed term is interpreted as zero. Since (\sum_i x_i=1),
[ x_1+x_3+x_5+\cdots=\frac{1+D}{2}. \tag{3} ]
2. Xiang's upper-bound strategy
Suppose Liu's marks initially divide the stick into (k\le n+1) pieces with lengths
[ a_1,a_2,\ldots,a_k, \qquad \sum_{i=1}^k a_i=1. ]
Case 1: (k\le n)
Xiang bisects every one of the (k) pieces. This uses at most (n) marks and produces equal pairs only.
During the claiming stage, Xiang can ensure that he receives one member of every equal pair: when Liu takes one member of a previously untouched pair, Xiang takes its mate; if Xiang ever has to move without such a response being required, he may take one member of an untouched pair himself. Consequently Liu receives at most one half of the total length, which is certainly no more than ((1+\delta)/2).
Case 2: (k=n+1)
Consider the (2^k) subset sums
[ \sum_{i\in I}a_i, \qquad I\subseteq{1,2,\ldots,k}. ]
They all lie in ([0,1]). After arranging them in nondecreasing order, two consecutive subset sums differ by at most
[ \frac1{2^k-1}=\frac1{2^{n+1}-1}=\delta. ]
Choose two distinct subsets giving such consecutive sums, and remove their common indices. We obtain disjoint subsets (P,Q) satisfying
[ 0\le \sum_{i\in P}a_i- \sum_{j\in Q}a_j \le\delta. \tag{4} ]
We now refine the pieces indexed by (P) and (Q). Repeatedly choose one currently unpaired piece from each side.
- If the two lengths are equal, pair them.
- If one is longer, cut the longer one so that one of the new parts has the length of the shorter piece; pair those two equal pieces and return the remainder to its side.
If (r=|P|+|Q|), this procedure uses at most (r-1) cuts: every cut completely removes at least one currently active piece from the comparison process. When one side is exhausted, the total length of all unpaired remainders is exactly
[ R= \sum_{i\in P}a_i- \sum_{j\in Q}a_j \le\delta. \tag{5} ]
Xiang bisects each of the other (k-r) original pieces, producing equal pairs. The total number of cuts is at most
[ (r-1)+(k-r)=k-1=n. ]
Therefore, except for a collection of pieces of total length (R\le\delta), all final pieces can be grouped into equal pairs.
In the claiming stage Xiang can secure one member of every equal pair. Liu can receive at most the other half of the paired mass and, in the worst case, all residual pieces. Hence Liu's total is at most
[ \frac{1-R}{2}+R = rac{1+R}{2} \le\frac{1+\delta}{2}. \tag{6} ]
Thus Xiang can always hold Liu to at most
[ \frac{1+\delta}{2} = rac{2^n}{2^{n+1}-1}. \tag{7} ]
3. Liu's matching lower-bound strategy
Liu uses exactly (n) marks to create (n+1) pieces with lengths
[ \delta,2\delta,4\delta,\ldots,2^n\delta. \tag{8} ]
It is enough first to work at the unnormalised scale, where the initial lengths are
[ 1,2,4,\ldots,2^n. \tag{9} ]
After Xiang makes at most (n) additional cuts, there are at most (2n+1) final pieces. Arrange their lengths as
[ x_1\ge x_2\ge\cdots\ge x_m. ]
We prove that their alternating gap satisfies
[ D=x_1-x_2+x_3-x_4+\cdots\ge1. \tag{10} ]
Each final piece originates from exactly one of the initial pieces (2^0,2^1,\ldots,2^n). Pair the sorted final pieces as
[ (x_1,x_2),(x_3,x_4),\ldots. ]
If (m) is odd, append one dummy piece of length zero and pair it with (x_m).
Construct a multigraph as follows.
- There is one vertex for every initial piece (2^i).
- Every pair of final pieces gives an edge joining the vertices corresponding to their origins.
- In the odd case, add one dummy vertex for the zero piece.
If (m) is even, the graph has (n+1) vertices and
[ \frac m2\le n ]
edges. If (m) is odd, the graph has (n+2) vertices and
[ \frac{m+1}{2}\le n+1 ]
edges. In either case the number of edges is smaller than the number of vertices. Hence at least one connected component is a tree.
Choose such a tree component and two-colour it. Assign a coefficient
[ \lambda_i\in{-1,1} ]
to its original vertices according to the two colours, and assign (\lambda_i=0) to every original vertex outside the component. If the dummy vertex lies in the component, colour it as well, but give it weight zero.
For every paired edge with endpoints (i,j),
[ \lambda_i+ \lambda_j=0. \tag{11} ]
Now consider the integer
[ S=\sum_{i=0}^{n}\lambda_i2^i. \tag{12} ]
It is nonzero. Indeed, if (j) is the largest index with (\lambda_j\ne0), then
[ |S| \ge2^j- \sum_{i=0}^{j-1}2^i =1. \tag{13} ]
On the other hand, the final pieces originating from the (i)-th initial piece have total length (2^i). Therefore (S) is also the signed sum of all final pieces, using the coefficient of their origin. Grouping this signed sum by the adjacent pairs and using (11), each pair contributes either zero or
[ \pm(x_{2r-1}-x_{2r}). ]
Consequently,
[ |S| \le (x_1-x_2)+(x_3-x_4)+\cdots =D. \tag{14} ]
Equations (13) and (14) prove (10).
Returning to the normalised scale multiplies every length by (\delta), so the alternating gap is at least (\delta). By the longest-piece strategy and (3), Liu can guarantee at least
[ \frac{1+\delta}{2} = rac{2^n}{2^{n+1}-1}. \tag{15} ]
The upper and lower bounds coincide. Therefore
[ \boxed{c_n=\frac{2^n}{2^{n+1}-1}}. ]